What formula to chose a nonlinear formula? Suppose I have a formula 
$$f(x) = x,$$ 
where $0 \leq x \leq 255$
Now I want to have a formula 
$$f(x) = y,$$ 
where $0 \leq x \leq 255$, where $f(0) = 0$, $f(255) = 255$ and e.g., $f(128) = 150$ (the value of $150$ might vary).
All other values should be interpolated.
So actually I want a function that is nonlinear (starts with $0$ and goes up to $255$, starting increasing fast and finishes increasing slow); opposite as a parabolic function.
What (kind of) function should I use?
 A: WA gives $$\frac{10933x-11x^2}{8128}$$
https://www.wolframalpha.com/input/?i=interpolate+((0,0),(128,150)(255,255))
A: I think one classical example of nonlinearity could be the gamma for color correction.

Instead of using the interval $X=0\ ..\ 255$ it is easier to work in $[0,1]$ by scaling $x=\frac X{255}$.
Now, remark that any $x^\gamma$ stays in $[0,1]$ since $0^\gamma=0$ and $1^\gamma=1$.
Remark: this is not exactly color correction, which is $x^{1/\gamma}$, but this is just a convention.

So you have linearity for $\gamma=1$ and non-linearity for any other value of the exponent.
Note that if you really want to work with bytes the result can be scaled back using $255\times\left(\frac X{255}\right)^\gamma$
Depending of your choice of $\gamma<1$ or $\gamma>1$ you will get either a fast increase at the start of the interval or at the end of the interval.
Try it here (move the g cursor): https://www.desmos.com/calculator/cnc2diykyx
A: To satisfy the requirements:
$$f(128)=150, f(255)=255$$
However, $f(0)=0$ can't be satisfied by this method. I assumed that f(1)=1, however 
you can change the numbers but not use zero, otherwise, there would be no inverse.

An example of the curve looks like this Curve
I can provide more info about the derivation if you want.
